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nexus9112 [7]
3 years ago
14

Where is the deepest part of the ocean

Physics
1 answer:
Mariana [72]3 years ago
5 0

Answer:

Trenches are the deepest part of the oceans, narrow and long depressions on the sea floor. One good example is the Mariana Trench, the deepest part of the ocean

Explanation:

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What are the advantages of nuclear energy ??​
Morgarella [4.7K]

Answer:

Welllllll

Explanation:

It is at a relativley low cost. It also has high energy density and it is throium. We have more than 80 years worth of uranium.

3 0
4 years ago
A negative charge of -2.5 C and a positive charge of 2.0 C are separated by 100 m. What is the force between the two charges?
Annette [7]

Explanation:

F = k |q1| |q2| / r^2

k = 9 * 10^9

q1 = - 2.5 C

q2 = 2 C

r = 100

r^2 = (10^2)^2 = 10^4

F = (9*10^9) * ( 2.5 ) ( 2) / ( 100)^2

F = 45* 10^9 / 10^4

F = 45 * 10^9 * 10 ^ -4 = 45 * 10^5 N

F = 45 * 10 ^ 5 N

6 0
3 years ago
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the
erik [133]

Answer:

Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).

Explanation:

5 0
3 years ago
During a warm-up, it's important to control your
Mariulka [41]

Answer:

breathing

Explanation:

8 0
4 years ago
A 0.20 kg mass attached to the end of a spring causes it to stretch 3.0 cm. What is the spring constant? What is the potential e
SVEN [57.7K]

Given that the mass is m = 0.2 kg and the displacement is x = 3 cm = 0.03 m

We have to find the spring constant and potential energy.

The spring constant can be calculated by the formula

\begin{gathered} F=\text{ kx} \\ mg\text{ = kx} \\ k\text{ = }\frac{mg}{x} \end{gathered}

Here, k is the spring constant.

g = 9.8 m/s^2 is the acceleration due to gravity.

Substituting the values, the spring constant will be

\begin{gathered} k=\frac{0.2\times9.8}{0.03} \\ =\text{ 65.33 N/m } \end{gathered}

The potential energy can be calculated as

\begin{gathered} U=\frac{1}{2}kx^2 \\ =\frac{1}{2}\times65.33\text{ }\times(0.03)^2 \\ =\text{ 0.0293 J} \end{gathered}

8 0
1 year ago
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